Answer
See the answer below.
Work Step by Step
a) $Ksp=[Ag^+][CN^-]=s^2$
$s=\sqrt{6\dot{}10^{-17}}=7.75\dot{}10^{-9}\ M$
b) $K=Ksp\dot{}Kf\rightarrow K=7.8\dot{}10^4$
Since the equilibrium constant is very large, the reaction is product favored, AgCN would dissolve in enough KCN.
c) $K=Ksp(AgCN)\dot{}Kf(Ag(S_2O_3)_2^{3-})$
$K=1.74\dot{}10^{-3}$
Assuming constant $[S_2O_3^{2-}]$:
$K=[Ag(S_2O_3)^{3-}][CN^-]/[S_2O_3^{2-}]^2$
$1.74\dot{}10^{-3}=s^2/0.10^2$
$s=4.18\dot{}10^{-3}\ M$
